Matrix of eigenvectors, relation to rotation matrix

1. Apr 12, 2013

Fluxthroughme

So I am given $B=\begin{array}{cc} 3 & 5 \\ 5 & 3 \end{array}$. I find the eigenvalues and eigenvectors: 8, -2, and (1, 1), (1, -1), respectively. I am then told to form the matrix of normalised eigenvectors, S, and I do, then to find $S^{-1}BS$, which, with $S = \frac{1}{\sqrt{2}}\begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array}$, I get $\begin{array}{cc} 8 & 0 \\ 0 & -2 \end{array}$. All great and dandy, that's the correct answer, and the text then asks me for the rotation matrix, which I give as $\begin{array}{cc} cos\theta & -sin\theta \\ sin\theta & cos\theta \end{array}$. However, when asked how the rotation matrix and S are related, I am clearly stumped; checking the answers, they have used $S = \frac{1}{\sqrt{2}}\begin{array}{cc} 1 & -1 \\ 1 & 1 \end{array}$, which I think is a valid choice, since the eigenvector could have been (1,-1) or (-1,1). But since I made a different choice, I cannot get the answer they are looking for (That S is the rotation matrix for theta is 45).

Thus, have I made a mistake in my calculation? In my reasoning? In my assumptions? Or is it just that I was unlucky to have picked the way I did and the question wasn't expecting that (I doubt the latter)?

2. Apr 12, 2013

vela

Staff Emeritus
You're just unlucky. The way you chose the eigenvectors, S is a rotation combined with a reflection across the line y=x. As you saw, you can choose the eigenvectors so that S is just a rotation.